# Null space of a hermitian matrix

I have the following matrix:

$\begin{pmatrix} 4 &2-i &-3i \\ 2+i &1 &1-i \\ 3i &1+i &1 \end{pmatrix}$

The following vector generates its null space:

$\begin{pmatrix} -1\\ 1+2i\\ 1 \end{pmatrix}$

When I tried to find the null space, I wrote the original system as a symmetric real matrix of size $6\times 6$ by separating my matrix into its real and imaginary components:

$\begin{pmatrix} 4 &2 &0 &0 &1 &3 \\ 2 &1 &1 &-1 &0 &1 \\ 0 &1 &1 &-3 &-1 &0 \\ 0 &-1 &-3 &4 &2 &0 \\ 1 &0 &1 &2 &1 &1 \\ 3 &1 &0 &0 &1 &1 \end{pmatrix}$

The vector $\begin{pmatrix} -1\\ 1\\ 1\\ 0\\ 2\\ 0 \end{pmatrix}$ is supposed to be in its nullspace but wolframalpha tells me it's not:

Wolfram tells me that $\left( \begin{array}{c} 0 \\ -2 \\ 0 \\ -1 \\ 1 \\ 1 \end{array} \right)$ is in the null space of the $6 \times 6$ matrix you've done there, which is similar to yours, but with the halves switched, and the wrong sign on the two.
Are you sure you have the right $6 \times 6$ matrix? Since the null space uses left-multiplication: $$\left( \begin{array}{ccc} 4 & 2-i & -3i \\ 2+i & 1 & 1-i \\ 3i & 1+i & 1 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right)$$ shouldn't the rows match your matrix and not the columns?
This would give you: $$\left( \begin{array}{cccccc} 4 & 2 & 0 & 0 & -1 & -3 \\ 2 & 1 & 1 & 1 & 0 & -1 \\ 0 & 1 & 1 & 3 & 1 & 0 \\ 0 & 1 & 3 & 4 & 2 & 0 \\ -1 & 0 & 1 & 2 & 1 & 1 \\ -3 & -1 & 0 & 0 & 1 & 1 \end{array} \right)$$ ...and, in fact, a quick bash into Wolfram reveals $\left( \begin{array}{c} -1 \\ 1 \\ 1 \\ 0 \\ 2 \\ 0 \end{array} \right)$ to be in the null space.